TEST1-solutions

# TEST1-solutions - TEST1 Tsoi(60250 This print-out should...

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– TEST1 – Tsoi – (60250) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A total charge Q = 28 . 7 μ C is deposited uniformly on the surface of a hollow sphere with radius R = 38 . 6 cm. Use ǫ 0 = 8 . 85419 × 10 12 C 2 / N m 2 . What is the magnitude of the electric field at the center of the sphere? 1. bardbl vector E 0 bardbl = 2. bardbl vector E 0 bardbl = 1 4 π ǫ 0 Q R 3. bardbl vector E 0 bardbl = 1 4 π ǫ 0 Q R 2 4. bardbl vector E 0 bardbl = 0 correct Explanation: Recall, by Gauss’ law, that a net electric flux exists only if there is a net charge within a closed surface. Mathematically, contintegraldisplay vector E · d vector A = q enclosed ǫ 0 Here, if we use a small area about the center of the sphere, there is no net charge enclosed. Hence, E 0 = 0. 002 (part 2 of 3) 10.0 points What is the magnitude of the electric field at a distance R 2 from the center of the sphere? 1. bardbl vector E R/ 2 bardbl = 1 4 π ǫ 0 4 Q R 2 2. bardbl vector E R/ 2 bardbl = 1 4 π ǫ 0 2 Q R 3. bardbl vector E R/ 2 bardbl = 1 4 π ǫ 0 4 Q R 4. bardbl vector E R/ 2 bardbl = 1 4 π ǫ 0 2 Q R 2 5. bardbl vector E R/ 2 bardbl = 0 correct Explanation: Now we wish to find E within the sphere at r = 10 cm. This is, in essence, the same situation as in part (a). No charge lies within a Gaussian sphere of radius r = 10 cm. Thus, again, E = 0. 003 (part 3 of 3) 10.0 points What is the magnitude of the electric field at a distance 77 . 2 cm from the center of the sphere? 1. 1001740.0 2. 169362.0 3. 515409.0 4. 520593.0 5. 432801.0 6. 257144.0 7. 359055.0 8. 404718.0 9. 339143.0 10. 623929.0 Correct answer: 4 . 32801 × 10 5 N / C. Explanation: Next, for a radius greater than the sphere’s radius, Gauss’ law yields a nontrivial answer contintegraldisplay vector E · d vector A = E (4 π r 2 ) = Q ǫ 0 or, E = Q 4 π ǫ 0 r 2 = 2 . 87 × 10 5 C 4 π ǫ 0 (0 . 772 m) 2 = 4 . 32801 × 10 5 N / C , where ǫ 0 = 8 . 85419 × 10 12 C 2 / N m 2 . 004 10.0 points A point charge q 1 is concentric with two spher- ical conducting thick shells, as shown in the figure below. The smaller spherical conduct- ing shell has a net charge of q 2 and the larger spherical conducting shell has a net charge of q 3 .

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– TEST1 – Tsoi – (60250) 2 q 3 q 2 q 1 R 1 R 2 R 3 R 4 R 5 r 1 r 2 r 3 r 4 Hint: Under static conditions, the charge on a conductor resides on the surface of the conductor. What is the charge Q r 3 on the inner surface of the larger spherical conducting shell? 1. Q r 3 = + q 1 2. Q r 3 = + q 1 q 2 3. Q r 3 = + q 1 + q 2 4. Q r 3 = q 1 5. Q r 3 = 0 6. Q r 3 = q 1 + q 2 7. Q r 3 = + q 1 + q 2 + q 3 8. Q r 3 = q 1 q 2 + q 3 9. Q r 3 = q 1 q 2 correct 10. Q r 3 = q 1 q 2 q 3 Explanation: The net charge inside a Gaussian surface located at r = R 4 must be zero, since the field in the conductor must be zero. Therefore, the charge on the inner surface of the large spherical conducting shell must be ( q 1 + q 2 ).
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